Little explorations with HP calculators (no Prime)

04072017, 08:52 PM
(This post was last modified: 04072017 08:53 PM by Gerson W. Barbosa.)
Post: #144




RE: Little explorations with the HP calculators
(04072017 06:00 PM)Han Wrote:(04072017 05:41 PM)Gerson W. Barbosa Wrote: Yes, we can! In fact that was my first approach. I've found cos Â = 1/4, cos B = 7/8 and cos Č = 11/16, the sides of the inner triangle being 2sqrt(5/8)x, sqrt(3/2)x and sqrt(47/8)x. Incidentally the area of the inner triangle = 7. The problem makes no requirement about the shape of the triangle . It only states that AE = CE, DB = 2*AD and CF = 3*BF. Provided these conditions are met, the area relationship between both triangles is preserved, no mater the shape. That's what the problem implies. I just trusted the problem statements and chose a convenient shape. I didn't know beforehand the inner triangle thus defined would be isosceles, but after calculating the lengths of the three sides this fact became evident and I took advantage of it to avoid the use of the more complicated Heron's formula. The invariance of the areas relationship regardless the shape needs indeed a proof, but that is not up to the solver, as the problem itself suggests it is true. If I needed a proof, the your solution might have sufficed. Although apparently only geometric solutions matter here, I think all approaches are valid as long they lead to the correct answer. 

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